Q.447. Elimination of θ from two equations.

Question 447.
Eliminate θ from (x/a) cos θ + (y/b) sin θ = 1 and xsin θ - y cos θ  = √(a² sin² θ + b² cos²θ).

Answer 447.
(x/a) cosθ + (y/b) sinθ = 1
=> (x/a)^2 cos^2 θ + 2(xy/ab) sinθ cosθ + (y/b)^2 sin^2 θ = 1
=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = sec^2 θ
=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = 1 + tan^2 θ
=> [(y/b)^2 - 1] tan^2 θ + 2(xy/ab) tanθ + [(x/a)^2 - 1] = 0
=> [(y^2 - b^2)/(x^2 - a^2)] * (a/b)^2 tan^2 θ + [2xy/(x^2 - a^2)] * (a/b) tanθ + 1 = 0 ... ( 1 )

xsinθ - ycosθ =√(a² sin² θ + b² cos² θ)
=> x^2 sin^2 θ - 2xysinθ cosθ + y^2 cos^2θ = a^2 sin^2 θ + b^2 cos^2 θ
=> x^2 tan^2 θ - 2xy tanθ + y^2 = a^2 tan^2 θ + b^2
=> (x^2 - a^2) tan^2 θ - 2xy tanθ + (y^2 - b^2) = 0
=> [(x^2 - a^2)/(y^2 - b^2)] tan^2 θ - [2xy / (y^2 - b^2)] tanθ + 1 = 0 ... ( 2 )

Comparing eqns. ( 1 ) and ( 2 ),

using coefficients of tanθ and constant terms,
[2xy/(x^2 - a^2)] * (a/b) / [- 2xy / (y^2 - b^2)] = 1
=> [1/(x^2 - a^2)] * (a/b) = - 1/(y^2 - b^2)
=> - a (y^2 - b^2) = b(x^2 - a^2)
=> bx^2 + ay^2 = ba^2 + ab^2
=> x^2/a + y^2/b = a + b.

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